I'm currently stuck on the following problem:
Let $\alpha$ be a 1-form on a connected 3-manifold $M$ such that $\alpha\wedge d\alpha$ is a volume form. Show that there exists a vector field $X$ on $M$ such that $i_X\alpha\equiv1$ and $i_X(d\alpha)\equiv0$.
I believe there are a few routes to solve this problem, but I keep getting stuck at each step. Working locally, I first wrote $$\alpha=\alpha_1dx+\alpha_2dy+\alpha_3dz$$ and noticed that $\alpha$ is non-vanishing. Indeed, if not, there exists a point $p\in M$ such that $\alpha_i(p)=0$ for $i=1,2,3$. Computing $\omega=\alpha\wedge d\alpha$, it follows that $\omega_p=0$, a contradiction. Thus, because $\alpha\neq0$, we get (by a theorem about $\alpha\wedge d\alpha$) that $\ker\alpha$ is not involutive, but I'm really not sure what I can do from here. I suppose that if we choose appropriate vector fields $X,Y\in\ker\alpha$, we do get that
$$d \alpha(X, Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X, Y])=-\alpha([X,Y])$$
is non-zero.
Alternatively, there is the usual expansion
$$i_X(\alpha\wedge d\alpha)=i_X\alpha\wedge d\alpha-\alpha\wedge i_Xd\alpha,$$
which connects both the conditions that $i_X\alpha\equiv1$ and $i_Xd\alpha\equiv0$ with the contraction of a volume form, but again I'm not able see where this leads me.
I'd really appreciate any help towards a full solution. I'm reviewing my knowledge of smooth manifold theory, so this problem is not homework (although it's from an old exam at my university). Thank you!
Best Answer
If you're not already aware, the vector field $X$ is called the Reeb vector field associated to $\alpha$.
The key piece of the construction is Jacobi's theorem, which states that the determinant of an odd-dimensional skew-symmetric matrix is zero. This implies that a skew 2-form such as $d\alpha$ has nonzero kernel when viewed as a map $T_pM\to T^*_pM$ given by $v\to\iota_vd\alpha$.
Since $\alpha\wedge d\alpha$ is a volume form, we must have that $\ker\alpha\cap\ker d\alpha=0$, and thus $\ker d\alpha$ is one dimensional and $\alpha|_{\ker d\alpha}$ is nonvanishing. The Reeb field $X$ is exactly the preimage of $\{1\}$ under $\alpha|_{\ker d\alpha}$